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Section15.3Bachet and Mordell Curves

Let's start by talking about \(y^3=x^2+2\) as a type of curve. Recall from Section 3.5 that Bachet de Méziriac first asserted this had one positive integer solution in 1621, very early in the development of modern number theory.

Example15.3.1

What is that solution?

Fermat, Wallis, and Euler also studied this and gave various discussions and proofs of this fact. As we saw earlier, this equation is actually one of a more general class of equations called the Mordell equation: \begin{equation*}y^3=x^2+k\; , \;\; k\in\mathbb{Z}\end{equation*} Louis Mordell, an American-born British mathematician, proved some remarkable theorems about this class of equations.

Notice that Mordell's set of curves are not quadratic/conic but rather cubic, which makes them more mysterious (and, as it happens, more useful for cryptography). There is a theorem that they can only have finitely many integer points (in fact, there are even useful bounds for how many that depend only on the prime factorization of \(k\)). At the same time, they are apparently “simple” enough that they can still have infinitely many rational points; Gerd Faltings won a Fields Medal for proving that higher-degree curves cannot.

Subsection15.3.1Verifying points don't exist

Proving things about Mordell's equation is quite tricky, but once in a while there is something you can do. For instance, we can verify something we can see in the interact above.

Proof

This is a simple version of a far more general statement.

Proof

There are lots of similar statements one can prove too. But there is a larger point, based on the very specific conditions on \(M\) and \(N\). Namely, if we want to prove anything about such equations with methods we currently have access to in this text, we have no hope of getting any general results.

Subsection15.3.2More on Mordell

Let's see what I mean by “no hope” here by returning to Bachet's original equation, \(y^3=x^2+2\). What are some naive things we can say?

  • It should be clear that \(x\) and \(y\) must have the same parity.

  • If they are both even then \(y^3\) is divisible by 4 but \(x^2+2\equiv 2\text{ (mod }4)\), which is impossible.

  • So \(x\) and \(y\) are both odd.

  • That doesn't really narrow things down too much, really.

Now, Euler nearly proves the following fact.

Proof

Where's the problem? It turns out you can say that a product which is a cube is a product of cubes in this situation, but it requires some (geometrically motivated) proof, just like with \(\mathbb{Z}[i]\). In his 1765 “Vollständige Anleitung zur Algebra”, sections 187-188 and 191, Euler explicitly says that this just works – in any number system with \(\mathbb{Z}[\sqrt{c}]\). He solves this one in section 193, and solves \(x^2+4=y^3\) using the same technique in section 192, without realizing the problem.

But we shouldn't be too hard on Euler! He was one of the first people to even consider some essentially random new number system of this type. And, in 1738, he gives a correct and full proof of the observation that \(8\) and \(9\) is the only time a perfect square is preceded by a perfect cube, which is Mordell's equation for \(k=-1\). (See also Question 3.5.1.)

If you are interested in more information about how to prove cases of Mordell's equation, there are many good resources, including a nice one on Keith Conrad's website.

In case you are wondering, even finding a bound on the size of the set of solutions to Mordell's equation for a given \(k\) is tricky.

  • Mordell, Siegel, and Thue all had a part after World War I in showing there are finitely many solutions for a given \(k\), but said nothing about how big \(x\) and \(y\) might be.

  • An early bound was that \begin{equation*}|x| < e^{10^{10}|k|^{10^4}}\end{equation*} which is of course ridiculously huge.

  • More recent conjectures are that \(x\) has absolute value less than \(e^C |d|^{2+\epsilon}\), where \(\epsilon\) is as small as you want and \(C\) seems to pretty close to one, probably less than two.

Finally, we have to mention a very famous result. Recall that these curves can have infinitely many rational points, even if they have finitely many (or zero) integer points. The following is a bit of a surprise, then; the rational points can still be described finitely.

If you like, the rational points might be infinite, but not too infinite.